Ch14 Chemical Equilibrium. Modified by Dr. Cheng-Yu Lai

Transcription

1 Ch14 Chemical Equilibrium Modified by Dr. Cheng-Yu Lai

2 CHEMICAL EQUILIBRIUM

3 Chemical Equilibrium: Chemical Equilibrium When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged Arrows going both directions ( ) indicates equilibrium in a chemical equation Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Eventually the rates are equal

4 Exp 1 Exp

5 Law of Mass Action For the reaction, when Rate for = Rate rev ja + kb lc + md K [ C] l [ A] [ D] j [ B] m k Where K is the equilibrium constant

6 Writing an Equilibrium Expression Write the equilibrium expression for the reaction: NO (g) NO(g) + O (g) K =??? [ NO] [ O K [ NO ] ]

7 Playing with Equilibrium Expressions NO (g) NO(g) + O (g) The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse NO(g) + O (g) NO (g) [ NO] [ O K [ NO ] ] K ' 1 K [ NO [ NO] ] [ O ]

8 Playing with Equilibrium Expressions NO (g) NO(g) + O (g) When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. NO (g) NO(g) + ½O (g) [ NO] [ O K [ NO ] ] K ' K 1 [ NO][ O ] [ NO ] 1

9 Playing with K If we multiply the equation by a constant, n nja + nkb nlc + nmd Then the equilibrium constant is K = [A] nj [B] nk = ([A] j [B] k ) n [C] nl [D] nm ([C] l [D] m ) n = Kn

10 Playing with K

11 Playing with K- Summary

12 Equilibrium Expressions Involving Pressure For the gas phase reaction: 3H (g) + N (g) NH 3 (g) K P ( P N P NH 3 )( P H 3 ) P, P 3 N, P NH H are equilibrium partial pressures K p K( RT ) n

13 Equilibrium and Pressure Kc and Kp SO (g) + O (g) SO 3 (g) Kp = (P SO3 ) (P SO ) (P O ) Kc = [SO 3 ] [SO ] [O ]

14 Equilibrium and Pressure Kc and Kp SO (g) + O (g) SO 3 (g) Kc = (P SO3 /RT) (P SO /RT) (P O /RT) Kc = (P SO3 ) (1/RT) (P SO ) (P O ) (1/RT) 3 Kc = Kp (1/RT) = Kp RT (1/RT) 3 Kp = kc (RT) -1

15

16

17 The units for K Are determined by the various powers and units of concentrations. They depend on the reaction.

18 Product Favored Equilibrium Large values for K signify the reaction is product favored When equilibrium is achieved, most reactant has been converted to product

19 Reactant Favored Equilibrium Small values for K signify the reaction is reactant favored When equilibrium is achieved, very little reactant has been converted to product

20 Solving for Equilibrium Concentration Consider this reaction at some temperature: H O(g) + CO(g) H (g) + CO (g) K =.0 Assume you start with 8 molecules of H O and 6 molecules of CO. How many molecules of H O, CO, H, and CO are present at equilibrium? Here, we learn about ICE the most important problem solving technique in the second semester. You will use it for the next 4 chapters!

21 Solving for Equilibrium Concentration H O(g) + CO(g) H (g) + CO (g) K =.0 Step #1: We write the law of mass action for the reaction:.0 [ H [ H ][ CO ] O][ CO]

22 Solving for Equilibrium Concentration Step #: We ICE the problem, beginning with the Initial concentrations H O(g) + CO(g) H (g) + CO (g) Initial: Change: Equilibrium: x -x +x +x 8-x 6-x x x

23 Solving for Equilibrium Concentration Step #3: We plug equilibrium concentrations into our equilibrium expression, and solve for x H O(g) + CO(g) H (g) + CO (g) Equilibrium: 8-x 6-x x x.0 (8 ( x)( x) x = 4 x)(6 x)

24 Solving for Equilibrium Concentration Step #4: Substitute x into our equilibrium concentrations to find the actual concentrations H O(g) + CO(g) H (g) + CO (g) Equilibrium: 8-x 6-x x x x = 4 Equilibrium: 8-4=4 6-4= 4 4

25 Example 14.5 Finding Equilibrium Constants from Experimental Concentration Measurements Consider the following reaction: A reaction mixture at 780 C initially contains [CO] = M and [H ] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant? Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 014 Pearson Education, Inc.

26 Example 14.6 Finding Equilibrium Constants from Experimental Concentration Measurements Consider the following reaction: A reaction mixture at 1700 C initially contains [CH 4 ] = M. At equilibrium, the mixture contains [C H ] = M. What is the value of the equilibrium constant? Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 014 Pearson Education, Inc.

27

28

29

30

31 Homogeneous Equilibria So far every example dealt with reactants and products where all were in the same phase. We can use K in terms of either concentration or pressure. Units depend on reaction.

32 Heterogeneous Equilibria If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn t change. As long as they are not used up they are not used up we can leave them out of the equilibrium expression. For example

33 For Example H (g) + I (s) K = [HI] [H ][I ] HI(g) But the concentration of I does not changefro example [H O]=15.6 M Combining [I ] into Kc, then K[I ]= [HI] = new K [H ]

34 Kp=Kc(RT)1

35

36

37 Le Chatelier s Principle

38 LeChatelier s Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress and restore a state of equilibrium. Henry Le Chatelier Please see the link below : entialchemistry/flash/flash.mhtml

39 Le Chatelier Translated: When you take something away from a system at equilibrium, the system shifts in such a way as to replace some what you ve taken away. When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you ve added.

40 Case 1

41 Case

42 The Effect of Volume Changes on Equilibrium When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules the reactant side. Right side of figure Pearson Education, Inc.

43 Case 4

44

45 The Reaction Quotient For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action. ja + kb lc + md Q [ C] l [ A] [ D] j [ B] m k

46 Significance of the Reaction Quotient If Q = K, the system is at equilibrium If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

47 Compare Q and K to predict the reaction Direction N O 4 (g) NO (g) K eq = 11 atm (T = 373 K) mix 0. mol of N O 4 with 0. mol of NO in a 4.0 L flask at 100 o C. Q = (P NO ) / P NO4 First find P NO and P NO4 using PV = nrt P NO = P NO4 = (0.0 mol)(0.081 L atm/mol K)(373 K)/(4.0 L) = 1.5 atm Q = (1.5 atm) /1.5 atm = 1.5 atm < 11 atm Q < K, so the reaction will proceed in the forward direction, N O 4 (g) --> NO (g) until it reaches equilibrium.